Question: Let $f(x) = x^{2}-8x-6$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Solution: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $x^{2}-8x-6 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = 1, b = -8, c = -6$ $ x = \dfrac{+ 8 \pm \sqrt{(-8)^{2} - 4 \cdot 1 \cdot -6}}{2 \cdot 1}$ $ x = \dfrac{8 \pm \sqrt{88}}{2}$ $ x = \dfrac{8 \pm 2\sqrt{22}}{2}$ $x =4 \pm \sqrt{22}$